A328407 G.f. A(x) satisfies: A(x) = A(x^2) + x * (1 + x) / (1 - x)^3.

1, 5, 9, 21, 25, 45, 49, 85, 81, 125, 121, 189, 169, 245, 225, 341, 289, 405, 361, 525, 441, 605, 529, 765, 625, 845, 729, 1029, 841, 1125, 961, 1365, 1089, 1445, 1225, 1701, 1369, 1805, 1521, 2125, 1681, 2205, 1849, 2541, 2025, 2645, 2209, 3069, 2401, 3125, 2601, 3549, 2809, 3645, 3025

Offset

1,2

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

G.f.: Sum_{k>=0} x^(2^k) * (1 + x^(2^k)) / (1 - x^(2^k))^3.

G.f.: (1/3) * Sum_{k>=1} J_2(2*k) * x^k / (1 - x^k), where J_2() is the Jordan function (A007434).

Dirichlet g.f.: zeta(s-2) / (1 - 2^(-s)).

a(2*n) = a(n) + 4*n^2, a(2*n+1) = (2*n + 1)^2.

a(n) = Sum_{d|n} A209229(n/d) * d^2.

Product_{n>=1} (1 + x^n)^a(n) = g.f. for A023871.

Sum_{k=1..n} a(k) ~ 8*n^3/21. - Vaclav Kotesovec, Oct 15 2019

Multiplicative with a(2^e) = (4^(e+1)-1)/3, and a(p^e) = p^(2*e) for an odd prime p. - Amiram Eldar, Oct 25 2020

Mathematica

nmax = 55; CoefficientList[Series[Sum[x^(2^k) (1 + x^(2^k))/(1 - x^(2^k))^3, {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x] // Rest
a[n_] := If[EvenQ[n], a[n/2] + n^2, n^2]; Table[a[n], {n, 1, 55}]
Table[DivisorSum[n, Boole[IntegerQ[Log[2, n/#]]] #^2 &], {n, 1, 55}]
f[p_, e_] := If[p == 2, (4^(e + 1) - 1)/3, p^(2*e)]; a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Array[a, 100] (* Amiram Eldar, Oct 25 2020 *)

Prog

(Magma) [n eq 1 select 1 else IsOdd(n) select n^2 else Self(n div 2)+n^2 :n in [1..55]]; // Marius A. Burtea, Oct 15 2019

Crossrefs

Cf. A000290, A001511, A007434, A016754, A023871, A129527, A209229, A328408.

Sequence in context: A246335 A255264 A321170 * A147407 A186297 A272986

Adjacent sequences: A328404 A328405 A328406 * A328408 A328409 A328410

Keyword

nonn,mult,easy

Author

Ilya Gutkovskiy, Oct 14 2019

Status

approved